新高考數(shù)學(xué)二輪復(fù)習(xí) 題型歸納演練專題3-5 利用導(dǎo)函數(shù)解決恒（能）成立問題原卷版

專題3-5利用導(dǎo)函數(shù)解決恒（能）成立問題目錄TOC\o"1-1"\h\u 1題型一：分離變量+最值法 1題型二：分類討論法 9題型三：同構(gòu)法 16題型四：最值定位法解決雙參不等式問題 23 32一、單選題 32二、多選題 38三、解答題 41題型一：分離變量+最值法【典例分析】例題1．（2023·全國·高三專題練習(xí)）若對任意的實(shí)數(shù)SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題2．（2022·全國·高三階段練習(xí)（文））設(shè)SKIPIF1<0是定義在SKIPIF1<0上的連續(xù)函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，且SKIPIF1<0.當(dāng)SKIPIF1<0時，不等式SKIPIF1<0恒成立，其中SKIPIF1<0為自然對數(shù)的底數(shù)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例題3．（2022·浙江·鎮(zhèn)海中學(xué)高二期中）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0在SKIPIF1<0上恒成立,求實(shí)數(shù)SKIPIF1<0的取值范圍;(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,求實(shí)數(shù)SKIPIF1<0的取值范圍.【提分秘籍】①若SKIPIF1<0)對SKIPIF1<0恒成立，則只需SKIPIF1<0；②若SKIPIF1<0對SKIPIF1<0恒成立，則只需SKIPIF1<0．③SKIPIF1<0，使得SKIPIF1<0能成立SKIPIF1<0SKIPIF1<0；④SKIPIF1<0，使得SKIPIF1<0能成立SKIPIF1<0SKIPIF1<0．【變式演練】1．（2022·甘肅省民樂縣第一中學(xué)高二期中（文））若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習(xí)）若不等式SKIPIF1<0對任意實(shí)數(shù)x都成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（多選）（2022·海南·模擬預(yù)測）若SKIPIF1<0時，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的值可以為（

）（附：SKIPIF1<0）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·湖北·仙桃市田家炳實(shí)驗(yàn)高級中學(xué)高三階段練習(xí)）若不等式SKIPIF1<0（其中SKIPIF1<0是自然對數(shù)的底數(shù)）對SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為________5．（2022·浙江寧波·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)若SKIPIF1<0在SKIPIF1<0上恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．6．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．（1）當(dāng)SKIPIF1<0時，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0，SKIPIF1<0（1）SKIPIF1<0處的切線方程；（2）若在區(qū)間SKIPIF1<0，SKIPIF1<0內(nèi)至少存在一個實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．題型二：分類討論法【典例分析】例題1．（2022·四川省岳池中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是自然對數(shù)的底數(shù)．(1)若SKIPIF1<0的最小值為0，求SKIPIF1<0；(2)若SKIPIF1<0在SKIPIF1<0上恒成立，求SKIPIF1<0的取值范圍．例題2．（2022·全國·高二專題練習(xí)）已知函數(shù)SKIPIF1<0的圖像在SKIPIF1<0處的切線與直線SKIPIF1<0垂直．(1)求SKIPIF1<0的解析式；(2)若SKIPIF1<0在SKIPIF1<0內(nèi)有兩個零點(diǎn)，求SKIPIF1<0的取值范圍；(3)若對任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的最大值．【提分秘籍】①首先可以把含參不等式整理成適當(dāng)形式如SKIPIF1<0、SKIPIF1<0等；②從研究函數(shù)的性質(zhì)入手，轉(zhuǎn)化為討論函數(shù)的單調(diào)性和極值或最值；③得出結(jié)論.【變式演練】1．（2023·陜西西安·高三期末（理））已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0，SKIPIF1<0，求實(shí)數(shù)a的取值范圍．2．（2022·江蘇·姜堰中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0恒成立，求正實(shí)數(shù)SKIPIF1<0的取值范圍.3．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0.（1）求函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間；（2）若存在SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍.題型三：同構(gòu)法【典例分析】例題1．（2022·河北·模擬預(yù)測）已知SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0恒大于0，求SKIPIF1<0的取值范圍.例題2．（2022·貴州·高三階段練習(xí)（理））已知SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍；(2)若不等式SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，求SKIPIF1<0的取值范圍．【提分秘籍】①對原不等式同解變形，如移項(xiàng)、通分、取對數(shù)、系數(shù)升指數(shù)等，把不等式轉(zhuǎn)化為左右兩邊是相同結(jié)構(gòu)的式子的結(jié)構(gòu)，根據(jù)“相同結(jié)構(gòu)”構(gòu)造輔助函數(shù)．②為了實(shí)現(xiàn)不等式兩邊“結(jié)構(gòu)”相同的目的，需時時對指對式進(jìn)行“改頭換面”，常用的方法有：SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，有時也需要對兩邊同時加、乘某式等.③SKIPIF1<0與SKIPIF1<0為常見同構(gòu)式：SKIPIF1<0，SKIPIF1<0；SKIPIF1<0與SKIPIF1<0為常見同構(gòu)式：SKIPIF1<0，SKIPIF1<0.【變式演練】1．（多選）（2022·云南·昆明一中高三階段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的可能的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·湖北·高三階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0，求SKIPIF1<0的取值范圍．3．（2022·江蘇蘇州·高三階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0時恒成立，求實(shí)數(shù)SKIPIF1<0的最小值.題型四：最值定位法解決雙參不等式問題【典例分析】例題1．（2022·湖南省臨澧縣第一中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0若對SKIPIF1<0,使得SKIPIF1<0成立,則實(shí)數(shù)SKIPIF1<0的最小值是SKIPIF1<0 B．SKIPIF1<0 C．2 D．3例題2．（2022·全國·高二課時練習(xí)）已知函數(shù)SKIPIF1<0,SKIPIF1<0,若對任意SKIPIF1<0都存在SKIPIF1<0使SKIPIF1<0成立,則實(shí)數(shù)SKIPIF1<0的取值范圍是______.例題3．（2022·江西·南昌十中高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若曲線SKIPIF1<0在SKIPIF1<0和SKIPIF1<0處的切線互相平行，求SKIPIF1<0的值；(2)求SKIPIF1<0的單調(diào)區(qū)間；(3)若對任意SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0，求SKIPIF1<0的取值范圍．【提分秘籍】最值定位法解決雙參不等式問題（1）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（3）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（4）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0【變式演練】1．（2022·廣東·汕頭市達(dá)濠華僑中學(xué)高三階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0.若對SKIPIF1<0，都SKIPIF1<0，使得不等式SKIPIF1<0成立，則SKIPIF1<0的最大值為（

）A．0 B．SKIPIF1<0 C．1 D．SKIPIF1<02．（2022·廣東·佛山市南海區(qū)九江中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若任意SKIPIF1<0，存在SKIPIF1<0，使SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是__________.3．（2022·全國·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若存在SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0成立，則實(shí)數(shù)a的取值范圍是___________.4．（2022·全國·高二課時練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，使得SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的最小值是_________．5．（2022·全國·高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍．一、單選題1．（2022·浙江·高二階段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0對任意的SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·廣東·紅嶺中學(xué)高二期中）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0，對SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·全國·高三專題練習(xí)）已知SKIPIF1<0，若?SKIPIF1<0，使SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0與SKIPIF1<0滿足：存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，則稱函數(shù)SKIPIF1<0為SKIPIF1<0的“友導(dǎo)”函數(shù)．已知函數(shù)SKIPIF1<0為函數(shù)SKIPIF1<0的“友導(dǎo)”函數(shù)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·廣東·高三開學(xué)考試）已知SKIPIF1<0，若對任意的SKIPIF1<0恒成立，則實(shí)數(shù)a的最小值為（

）A．e B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·安徽滁州·高二期末）已知當(dāng)SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2022·遼寧沈陽·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0使得SKIPIF1<0成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2022·河南·濮陽南樂一高高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0．若SKIPIF1<0，都SKIPIF1<0，使SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題9．（2022·江蘇·句容碧桂園學(xué)校高三階段練習(xí)）已知函數(shù)SKIPIF1<0，滿足對任意的SKIPIF1<0，SKIPIF1<0恒成立，則實(shí)數(shù)a的取值可以是（

）A．SKIPIF1<0